Question

The ratio of the sums of m and n terms of an A.P. is m²: n² . Show that the ratio of mth and nth term is (2m - 1): (2n – 1).​

Answer 1

Answer:

just see the answer

Step-by-step explanation:

S(m):S(n)=m²:n²

then let S(m) and S(n) be m²x and n²x respectively

then S(m-1) and S(n-1) will be (m-1)²x and (n-1)²x respectively

the T(m) will be S(m)-S(m-1) or m²x-(m-1)²x or (2m-1)x

similarly T(n) will be (2n-1)x

hence T(m):T(n) is (2m-1):(2n-1).

(proved)

Answer 2

Answer:

See the explanation

Step-by-step explanation:

Let a be the first term and d be the common difference of the given AP

A.T.Q

Sum of m terms / Sum of n terms  = m²/n²

⇒m/2[2a + (m-1)d] / n/2[2a +(n-1)d] = m²/n²

⇒2a + (m-1)d / 2a + (n-1)d = m/n

⇒n[2a + (m-1)d] = m[2a + (n-1)d]

⇒2na +nmd -nd = 2ma + nmd -md

⇒2na -nd = 2ma - md

⇒2na -2ma = nd - md

⇒2a(n-m) = (n-m)d

⇒2a =d ------------- (1)

Now , we know that nth term = a + (n-1)d

Therefore, mth term / nth term = a+(m-1)d / a+(n-1)d

                                                   = a+(m-1)2a / a+(n-1)2a -----------[From eq.1]

                                                   =a[1+(m-1)2] / a[1+(n-1)2]

                                                   = 1+2m -2 / 1+2m -2

                   mth term / nth term= 2m-1 / 2n-1

                            HENCE PROVED !!